Some relativity question(s). I am especially interested in what Einstein would say, but there are other theories.

Spaceship X is connected to spaceship Y (line ahead) by a tight elastic thread & spaceship O is nearby. X & Y & O are stationary. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....(A1) X & Y accelerate at the same rate & there is no observer on O which is stationary.(A2) X & Y accelerate at the same rate & there is an observer on O which is stationary.(A3) X & Y accelerate at the same rate & there is an observer on O which accelerates likewise.

O is stationary & X & Y are going past at hi speed connected by the tight thread. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....(B1) X & Y decelerate & there is no observer on O which is stationary.(B2) X & Y decelerate & there is an observer on O which is stationary.(B3) X & Y decelerate & there is an observer on O which accelerates at the same rate in the opposite direction.

Some relativity question(s). I am especially interested in what Einstein would say, but there are other theories.

Spaceship X is connected to spaceship Y (line ahead) by a tight elastic thread & spaceship O is nearby. X & Y & O are stationary. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....(A1) X & Y accelerate at the same rate & there is no observer on O which is stationary.(A2) X & Y accelerate at the same rate & there is an observer on O which is stationary.

Observers have no effect on what happens. Perhaps you are thinking of quantum mechanics where observers arguably play a role. So the first two situations are the same, and the string stretches or doesn’t, regardless of if anyone is watching.

If X and Y accelerate, the string stretches because force is being applied to each end to accelerate it, and the middle is going to sag a bit until the greater tension at the front and lower tension at the back can account for the acceleration of the mass of the string. That’s Newton’s answer, not Einstein’s.

Einstein says that acceleration itself has no effect, but as speed picks up, the string maintains its tension at all times, but its length contracts relative to the original frame of O as the speed of the two ships X and Y reach relativistic speeds.

The situation is little different from one long ship with X being the bottom and Y being the top, with a string between them running along the interior of the ship. That string will be in an identical state for the duration of the acceleration from the perspective of the ship, but relative to frame O, the ship and its string gets shorter as its speed becomes a significant fraction of light speed. This shortening has no effect on the tension of the string, as it gets actually shorter, not just less stretched.

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(A3) X & Y accelerate at the same rate & there is an observer on O which accelerates likewise....(B3) X & Y decelerate & there is an observer on O which accelerates at the same rate in the opposite direction.

If our observer is accelerating with X and Y, he sees zero change for the duration of the experiment. Frame O is not inertial in A3 or B3. There is no local way of distinguishing this from X, Y, and O all being stationary on a planet with gravity identical to the acceleration rate of the ships.

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O is stationary & X & Y are going past at hi speed connected by the tight thread. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....(B1) X & Y decelerate & there is no observer on O which is stationary.(B2) X & Y decelerate & there is an observer on O which is stationary.

This is just the reverse of above. The thread has identical tension the whole time in any frame, but relative to inertial frame O, its length becomes less dilated (shortened) as the thread speed is reduced.

As for the question of if the thread breaks: No. The tension is the same the whole time while accelerating, and even if it wasn't, it was described as an elastic thread, so one end could be tied to O as the ships drag the other end away, making it very long, but not broken. If it breaks beyond a length L, then L was not specified.

The situation is little different from one long ship with X being the bottom and Y being the top, with a string between them running along the interior of the ship.

Little different, but different nonetheless. That explanation almost works if X and Y are pretty close to each other, but in fact X and Y clocks will be dilated relative to each other. Y will run faster (being less deep in the equivalent gravity well) and will measure the same speed change as having taken place in more time, thus less acceleration, which violates the 'same acceleration' stipulation of the OP.

Another way of looking at it is to imagine one ship that is 10 light years in length, with Y at the top and X at the bottom at the 'origin location'. It accelerates in one day (as measured by O frame say) up to .866c. In that time, X is less than half a light day (let's say 10 light-hours) from its starting point, and the length of the ship is dilated to only 5 light years in length now, so Y must be only 5 light years away + 10 light hours. That's impossible. It puts Y closer to the 'origin' than when it started out. The answer I gave above should not have used the one-ship analogy.

For two ships accelerating like that, beginning simultaneously (in frame O) for one day, they both move ahead by that 10 light hours. The string is still 10 light years long in O's frame, but stretched now, since its new proper length is 20 light years.In the new frame 'P' of X and Y (.866c relative to O) where both those ships will be stationary after the acceleration: well, first of all, they have to stop accelerating and coast for over 17 years to be relatively stationary again. So they do that. X and Y, initially moving at .866c relative to P, with X in front, are only 5 light years apart, so the string's dilated length is 5 light years. Y accelerates to a halt first, then X over 17 years later. When they're done doing that, the string is stationary and 20 light years long now. It doesn't break because it is elastic, but if it wasn't elastic, yes it would break.

I will digest all of that later & after some more replies give my own ideas. I think that anyone can impose any extra conditions they like, but it might get very complicated.I am happy to assume that the string & spaceships are massless & that there is no nearby mass etc in that part of the cosmos (ie true zero gravity) -- i prefer to imagine X & Y to be in line horizontally, both facing to left, with X ahead (both going rt to left), & with O sitting on the centerline of the picture (at least initially) & well below X & Y.

Some relativity question(s). I am especially interested in what Einstein would say, but there are other theories.

Spaceship X is connected to spaceship Y (line ahead) by a tight elastic thread & spaceship O is nearby. X & Y & O are stationary. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....(A1) X & Y accelerate at the same rate & there is no observer on O which is stationary.(A2) X & Y accelerate at the same rate & there is an observer on O which is stationary.

Observers have no effect on what happens. Perhaps you are thinking of quantum mechanics where observers arguably play a role. So the first two situations are the same, and the string stretches or doesn’t, regardless of if anyone is watching. Comment. Ok, Einsteinians & aetherists would agree here.

If X and Y accelerate, the string stretches because force is being applied to each end to accelerate it, and the middle is going to sag a bit until the greater tension at the front and lower tension at the back can account for the acceleration of the mass of the string. That’s Newton’s answer, not Einstein’s. Comment. Ok, but i wasnt really looking for real-life effects, ie we would usually assume the string & spaceships to be massless.

Einstein says that acceleration itself has no effect, but as speed picks up, the string maintains its tension at all times, but its length contracts relative to the original frame of O as the speed of the two ships X and Y reach relativistic speeds. Comment. I think that Einstein would say that while the relative speed tween XY & O is increasing then observer O would see that the distance tween X & Y gradually shortens (because space contracts along that line)(& anything in that space contracts accordingly)(ie string shortens, & X & Y each shorten), but, i think that heshe would say that there is no stretching or slackening of the string, it maintains its tension at all times (so u & i agree here). However observer X & observer Y would say that the distance tween X & Y doesnt change, & they would say that there is no stretching or slackening of the string in agreement with observer O but for a different reason.

Me myself (ie an aetherist)(ie a neo-Lorentzian)(ie an anti-Einsteinian), i believe that an aetherwind blows throo the spaceships, & i believe that all objects shorten (or lengthen) depending on the change in the apparent aetherwind. Therefore before voicing an opinion re the fate of the string in A1 & A2 i would firstly need to know the speed & direction of the (initial) aetherwind relative to X~Y~O (when stationary), plus i would need to know the magnitude & direction of the subsequent acceleration of X~Y. If the acceleration resulted in a stronger apparent aetherwind felt by X & Y measured along the line of X~Y then (a) X would shorten & Y would shorten, which would result in a wider gap tween X & Y, & (b) the string would shorten, which together with the widening of the gap would mean that the string would need to stretch (& might break). But if the acceleration resulted in a weaker aetherwind then X & Y would each lengthen, the gap tween would shorten, & the string would itself lengthen, & the string would slacken. If the initial aetherwind was a tailwind then early on the string would slacken & then later the string would stretch.

That describes what must happen, but my original question is moreso re what do observers XYO see happen. Observer O is stationary throughout & hencely shehe will see the string stretching or slackening precisely as it happens. The thing is that in accordance with aether theory all observers will see the actual length & size & shape of all objects at all times, the magnitude & direction of the aetherwind blowing throo the observer or blowing throo the object makes no difference. This is because any change in size or shape of the object is matched by the change in the size & shape of the observer's eye(s).

Aetherists believe that an objects proper true length & size & shape is the length & size & shape that it enjoys when in the absolute reference frame (sometimes called preferred frame). In the ARF the aetherwind is zero kmps, hencely there is no contraction, an object's length is its true length. Now the strange thing is that all observers will see an object's true length at all times, as long as the observer & the object have the same velocity (eg if both appear to be stationary out in space). But if there is a relative velocity tween observer & object then it is highly unlikely that the observer will be seeing the object's true length (because the distortion of the eyes doesnt match the distortion of the object).

There is little need to look at B123, the above aetheristic rules can be applied to tell whether the string stretches or slackens or no change (or whether the string does one thing for a while & then the other).

In a sense similar things can happen in the Einsteinian universe, eg if X & Y are at first getting closer, then become level with O, & then fly away, then the relative velocity relative to O will be decreasing then zero then increasing, but in all 3 cases the Einsteinian string will not change its Einsteinian tension.

The situation is little different from one long ship with X being the bottom and Y being the top, with a string between them running along the interior of the ship. That string will be in an identical state for the duration of the acceleration from the perspective of the ship, but relative to frame O, the ship and its string gets shorter as its speed becomes a significant fraction of light speed. This shortening has no effect on the tension of the string, as it gets actually shorter, not just less stretched.

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(A3) X & Y accelerate at the same rate & there is an observer on O which accelerates likewise....(B3) X & Y decelerate & there is an observer on O which accelerates at the same rate in the opposite direction.

If our observer is accelerating with X and Y, he sees zero change for the duration of the experiment. Frame O is not inertial in A3 or B3. There is no local way of distinguishing this from X, Y, and O all being stationary on a planet with gravity identical to the acceleration rate of the ships. Comment.

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O is stationary & X & Y are going past at hi speed connected by the tight thread. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....(B1) X & Y decelerate & there is no observer on O which is stationary.(B2) X & Y decelerate & there is an observer on O which is stationary.

This is just the reverse of above. The thread has identical tension the whole time in any frame, but relative to inertial frame O, its length becomes less dilated (shortened) as the thread speed is reduced. Comment.

As for the question of if the thread breaks: No. The tension is the same the whole time while accelerating, and even if it wasn't, it was described as an elastic thread, so one end could be tied to O as the ships drag the other end away, making it very long, but not broken. If it breaks beyond a length L, then L was not specified. Comment.

Ok, but i wasnt really looking for real-life effects, ie we would usually assume the string & spaceships to be massless.

Fair enough, unless you start drawing conclusions from such impossibilities. If I tug on a long nearly-massless string, the other end will not move right away, subject to the speed of sound. Likewise the top of any rocket cannot move right away when the bottom starts to move. The sting is going to stretch, massless or no. We can ignore that, but keep it in mind.Meanwhile, I consider the two ships to be points. It just complicates things to give them length. It's the point where the string is attached that matters, not the rest.

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I think that Einstein would say that while the relative speed tween XY & O is increasing then observer O would see that the distance tween X & Y gradually shortens (because space contracts along that line)(& anything in that space contracts accordingly)(ie string shortens, & X & Y each shorten), but, i think that heshe would say that there is no stretching or slackening of the string, it maintains its tension at all times (so u & i agree here).

Well I did say that the post to which you are responding was wrong. I assumed incorrectly that X and Y were opposite ends of a ship that had thrusters all along its length, not just at the bottom, with the string running the length inside. Read my 2nd reply. The string stretches, and would break if not elastic.

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Me myself (ie an aetherist)(ie a neo-Lorentzian)(ie an anti-Einsteinian),

Gotcha. I suppose that's why you're posting in this forum and not the main one. I should have clued off from your user name when I first replied.

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Therefore before voicing an opinion re the fate of the string in A1 & A2 i would firstly need to know the speed & direction of the (initial) aetherwind relative to X~Y~O (when stationary),

You said all objects were stationary at the beginning. Isn't that definition enough for you?The B examples had O stationary and X and Y moving. You were quite clear about that, except in B3 where there really isn't anything that stands still.

Disclaimer: I read B3 wrong. I didn't see that O was accelerating in the opposite direction. I took it to work like A3 where they all went the same way. My answer for B3 was pretty much identical to A3 for that reason.

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If the acceleration resulted in a stronger apparent aetherwind felt by X & Y measured along the line of X~Y then (a) X would shorten & Y would shorten, which would result in a wider gap tween X & Y, & (b) the string would shorten, which together with the widening of the gap would mean that the string would need to stretch (& might break). But if the acceleration resulted in a weaker aetherwind then X & Y would each lengthen, the gap tween would shorten, & the string would itself lengthen, & the string would slacken.

It more depends on if X or Y starts accelerating first, or more specifically, in which frame they 'simultaneously' start their identical changes of velocity. If you're going to have an absolute reference, you need to specify it in your question. In the B examples, it seems to be the frame of X and Y after they finish.

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That describes what must happen, but my original question is moreso re what do observers XYO see happen. Observer O is stationary throughout & hencely shehe will see the string stretching or slackening precisely as it happens.

No, that cannot be. Observer O is not in the presence of either X or Y, so does not see any of their events as they happen, but only later when light reaches O from those events. O can compute the length of the string, but does not directly observe it 'as it happens'. O can do that same computation by simply consulting the flight schedule of X and Y and need not observe them at all.

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Aetherists believe that an objects proper true length & size & shape is the length & size & shape that it enjoys when in the absolute reference frame (sometimes called preferred frame).

I am familiar with the view.

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In a sense similar things can happen in the Einsteinian universe, eg if X & Y are at first getting closer, then become level with O, & then fly away, then the relative velocity relative to O will be decreasing then zero then increasing, but in all 3 cases the Einsteinian string will not change its Einsteinian tension.

The relative velocity remains constant in that scenario. You mention no acceleration. It seems to be a description of a fly-by. Perhaps I misread it and you mean something comes in, slowing to a stop, then returning the way it came, in which case, only one of X and Y can actually stop at O, so the other is never 'level with O', and the string will change tension with speed, breaking if it can stretch only so far.If your aether theory doesn't predict that, then it is wrong. But it does predict that.

Ok, but i wasnt really looking for real-life effects, ie we would usually assume the string & spaceships to be massless.

Fair enough, unless you start drawing conclusions from such impossibilities. If I tug on a long nearly-massless string, the other end will not move right away, subject to the speed of sound. Likewise the top of any rocket cannot move right away when the bottom starts to move. The sting is going to stretch, massless or no. We can ignore that, but keep it in mind. Comment. Ok. Meanwhile, I consider the two ships to be points. It just complicates things to give them length. It's the point where the string is attached that matters, not the rest. Comment. I think that the shortening of spaceships is critical (or can be), in the aetheric universe it magnifies any stretching or slackening of the string. All objects (spaceship)(string) shorten or lengthen equally (in any given direction)(if the aetherwind changes in that direction). But in the aetheric universe the gap always does the opposite, when objects shrink the gap widens (in any given direction), hencely its a double whammy. Meanwhile over in the Einsteinian universe Einsteinians cannot agree amongst themselves. Most of them say that a change in relative velocity contracts or expands the space in that direction (as seen by the stationary observer)(meaning all observers)(because all observers are stationary from their own point of view), & any object sitting or moving or accelerating in that space contracts or expands with that space. In which case the string never stretches or slackens & never breaks.The additional problem in the Einsteinian universe being that nothing need be real, every relativity question is solved using Einstein's gamma which is a math-trick-model that is designed to give goodish answers, reality is not needed. The golden example of this is of course the Twin Paradox (i dont want to start a new argument here), which involves the gamma for ticking not length, but paradoxes involving length are no less serious, but are not as breathtaking (eg the spaceships~thread paradox). The answer is that Einsteinian paradoxes are not real paradoxes, because Einsteinian relativity situations are not real, they are math-trick-models designed to give a quick & sometimes good answer. I am ok with math-models & math-trick-models, but in the modern precision era Einsteinian relativity is found to be missing the target, & must be replaced by a better model, which will involve aether.

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I think that Einstein would say that while the relative speed tween XY & O is increasing then observer O would see that the distance tween X & Y gradually shortens (because space contracts along that line)(& anything in that space contracts accordingly)(ie string shortens, & X & Y each shorten), but, i think that heshe would say that there is no stretching or slackening of the string, it maintains its tension at all times (so u & i agree here).

Well I did say that the post to which you are responding was wrong. I assumed incorrectly that X and Y were opposite ends of a ship that had thrusters all along its length, not just at the bottom, with the string running the length inside. Read my 2nd reply. The string stretches, and would break if not elastic. Comment. I will have a look. But offhand i am puzzled at why the string might want to break, it should not want to break (ie stretch) from the viewpoint of any & all Einsteinian observers. I will have a look.

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Me myself (ie an aetherist)(ie a neo-Lorentzian)(ie an anti-Einsteinian),

Gotcha. I suppose that's why you're posting in this forum and not the main one. I should have clued off from your user name when I first replied. Comment. Yes my comments are likely to involve a dynamic aether, non-mainstream.

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Therefore before voicing an opinion re the fate of the string in A1 & A2 i would firstly need to know the speed & direction of the (initial) aetherwind relative to X~Y~O (when stationary),

You said all objects were stationary at the beginning. Isn't that definition enough for you? Comment. No, i dont think that it is enough, i need to know the aetherwind blowing throo the stationary observers, because the aetherwind will have a projection~value along the X~Y line, & that initial condition is critical, eg it might initially be a headwind or a tailwind or (unlikely) zero. Perhaps i could have said that at minimum i need to know the value of the projection of the aetherwind along the X~Y line -- but i think that that is an error -- i need to know the full value of the full wind (the explanation is not long but i will give it a miss for now).The B examples had O stationary and X and Y moving. You were quite clear about that, except in B3 where there really isn't anything that stands still. Comment. Yes, in B3 O is initially still but as u say doesnt stay still.

Disclaimer: I read B3 wrong. I didn't see that O was accelerating in the opposite direction. Comment. Yes i think i saw that. I took it to work like A3 where they all went the same way. My answer for B3 was pretty much identical to A3 for that reason. Comment.

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If the acceleration resulted in a stronger apparent aetherwind felt by X & Y measured along the line of X~Y then (a) X would shorten & Y would shorten, which would result in a wider gap tween X & Y, & (b) the string would shorten, which together with the widening of the gap would mean that the string would need to stretch (& might break). But if the acceleration resulted in a weaker aetherwind then X & Y would each lengthen, the gap tween would shorten, & the string would itself lengthen, & the string would slacken.

It more depends on if X or Y starts accelerating first, or more specifically, in which frame they 'simultaneously' start their identical changes of velocity. If you're going to have an absolute reference, you need to specify it in your question. In the B examples, it seems to be the frame of X and Y after they finish. Comment. I like your comment re simultaneity, i hadnt thort of that. Simultaneity can be a problem in an Einsteinian universe, but it aint a major problem in the aether universe. In the aether universe the present instant of time is universal, & there is no such thing as time dilation, it is merely a ticking dilation. Either way the best way of ignoring the problems of ticking & simultaneity etc is i think to simply say that for the purposes of observation the speed of light is infinite.

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That describes what must happen, but my original question is moreso re what do observers XYO see happen. Observer O is stationary throughout & hencely shehe will see the string stretching or slackening precisely as it happens.

No, that cannot be. Observer O is not in the presence of either X or Y, so does not see any of their events as they happen, but only later when light reaches O from those events. O can compute the length of the string, but does not directly observe it 'as it happens'. O can do that same computation by simply consulting the flight schedule of X and Y and need not observe them at all. Comment. Good points. This is solved by assuming that observational light travels at infinite speed.We see the same problem in Einstein's train & lightning thortX. Einstein assumes that (1) observational light travels at infinite speed (but doesn't say so) & (2) assumes that an observer can see a ray of light (this would need an assistant with a smoke machine).

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Aetherists believe that an objects proper true length & size & shape is the length & size & shape that it enjoys when in the absolute reference frame (sometimes called preferred frame).

I am familiar with the view.

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In a sense similar things can happen in the Einsteinian universe, eg if X & Y are at first getting closer, then become level with O, & then fly away, then the relative velocity relative to O will be decreasing then zero then increasing, but in all 3 cases the Einsteinian string will not change its Einsteinian tension.

The relative velocity remains constant in that scenario. You mention no acceleration. It seems to be a description of a fly-by. Perhaps I misread it and you mean something comes in, slowing to a stop, then returning the way it came, in which case, only one of X and Y can actually stop at O, so the other is never 'level with O', and the string will change tension with speed, breaking if it can stretch only so far. Comment. True, in the Einsteinian universe we would need to recognize that as the array of X~string~Y passes only one point of the array can be directly level or opposite O at any one instant, & i imagine that this can result in some complications when a half of the string is coming while a half is going (from O's view). However this doesnt make any difference in the aether universe.If your aether theory doesn't predict that, then it is wrong. But it does predict that. Comment.

All objects (spaceship)(string) shorten or lengthen equally (in any given direction)(if the aetherwind changes in that direction).

Your theory has the aetherwind changing? Thatís not the neo-Lortentzian view.

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But in the aetheric universe the gap always does the opposite, when objects shrink the gap widens (in any given direction), hencely its a double whammy.

The gap changes only in the dimension of movement. So if X and Y are side-by side, string or not between them, the gap between them and the string length remains the same as they accelerate in parallel. The ships get shorter, but no less wide. The string length is unaltered but it gets thinner in one dimension, squashing to sort of an elliptical cross section. Both Einstein and Lorentz claim this, but apparently not you, so your view contradicts empirical physics.

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Meanwhile over in the Einsteinian universe Einsteinians cannot agree amongst themselves. Most of them say that a change in relative velocity contracts or expands the space in that direction (as seen by the stationary observer)(meaning all observers)(because all observers are stationary from their own point of view), & any object sitting or moving or accelerating in that space contracts or expands with that space. In which case the string never stretches or slackens & never breaks.

If anybody says any of that, they donít know their physics. If you want to call them Einsteinians, fine, but you seem to be talking about uneducated people.

What you say about a string not breaking is true where X and Y are different ends of the same ship, accelerating as one unit, not two separate units. This is exactly equivalent to a string hanging in a tall building with Y at the top and X at the bottom, with gravity at Y equal to the acceleration of Y. The string just sits there, not breaking.

OK. I sort of figured that out above where you talked about the wind changing. Let me know how that works out for you. Do fast moving things contract in length? I was going with the Lortentz view on this, but youíve got different ideas, so I have no clue as to what your rules are. Length contraction would seem to be in direct contradiction with a changing etherwind.

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No, i dont think that it is enough, i need to know the aetherwind blowing throo the stationary observers

How is stationary defined if not relative to the aetherwind? Minkowski spacetime has inertial reference frames, but the typical Lortentz interpretation says only one of those reference frames is preferred: the one where the aetherwind doesnít blow.

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In the aether universe the present instant of time is universal

Somehow I guessed that you would also adopt this divergence from Lorentzís ideas.

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Good points. This is solved by assuming that observational light travels at infinite speed.

Now youíre living in another universe. I can assume no such thing. You can disprove any bit of physics you like with this assumption that directly contradicts empirical measurements.

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We see the same problem in Einstein's train & lightning thortX. Einstein assumes that (1) observational light travels at infinite speed (but doesn't say so) & (2) assumes that an observer can see a ray of light (this would need an assistant with a smoke machine).

Nonsense. Youíre making up fiction. The whole experiment revolves around finite but constant light speed. Nobody is observing rays. Light reaches a detector and is measured but once.

All objects (spaceship)(string) shorten or lengthen equally (in any given direction)(if the aetherwind changes in that direction).

Your theory has the aetherwind changing? That’s not the neo-Lorentzian view. Comment. The background aetherwind blowing at 500 kmps south to north at 20 deg off Earth's spin-axis might not change much, but the apparent wind blowing throo u & me gradually changes during each day & year due to Earth's 0.4 kmps spin & 30 kmps orbit. For example Demjanov's 1970 MMX in Obninsk showed that the horizontal projection of the wind was 140 kmps at one time of day & had a max of 480 kmps at another time of day (MMXs are done horizontally). And the apparent wind will be different if u are driving at speed. I say apparent, but of course it is not apparent to u or me, but can be if we have the right sort of detector. This is all in accord with neo-Lorentz, apart from minor disagreement re exact numbers.

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But in the aetheric universe the gap always does the opposite, when objects shrink the gap widens (in any given direction), hencely its a double whammy.

The gap changes only in the dimension of movement. [Yes] So if X and Y are side-by side, string or not between them, the gap between them and the string length remains the same as they accelerate in parallel. The ships get shorter, but no less wide. The string length is unaltered but it gets thinner in one dimension, squashing to sort of an elliptical cross section. Both Einstein and Lorentz claim this, but apparently not you, so your view contradicts empirical physics. Comment. No i said that the spaceships were in line (ie line astern)(not side by side).

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Meanwhile over in the Einsteinian universe Einsteinians cannot agree amongst themselves. Most of them say that a change in relative velocity contracts or expands the space in that direction (as seen by the stationary observer)(meaning all observers)(because all observers are stationary from their own point of view), & any object sitting or moving or accelerating in that space contracts or expands with that space. In which case the string never stretches or slackens & never breaks.

If anybody says any of that, they don’t know their physics. If you want to call them Einsteinians, fine, but you seem to be talking about uneducated people. Comment. I was referring to the rezults of a poll at CERN – some said that the string doesn’t break, some said that it breaks (last i heard all of them were still getting paid).

What you say about a string not breaking is true where X and Y are different ends of the same ship, accelerating as one unit, not two separate units. This is exactly equivalent to a string hanging in a tall building with Y at the top and X at the bottom, with gravity at Y equal to the acceleration of Y. The string just sits there, not breaking. Comment. In my string-thort-X X & Y are allways doing exactly the same thing (ie at the same time). I don’t know about that tall building stuff, my thort-X is out in space well away from mass etc.

OK. I sort of figured that out above where you talked about the wind changing. Let me know how that works out for you. Do fast moving things contract in length? I was going with the Lortentz view on this, but you’ve got different ideas, so I have no clue as to what your rules are. Length contraction would seem to be in direct contradiction with a changing etherwind. Comment. Fast things might be contracted or might not be. That’s why i said I needed to know the exact details of the background aetherwind. If the fast thing is going against the background wind then it will be contracted, if with then it will be longer. If almost side-on to the wind (& slightly with)(say at 91 deg) then the length might be non-changed.

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No, i dont think that it is enough, i need to know the aetherwind blowing throo the stationary observers

How is stationary defined if not relative to the aetherwind? Minkowski spacetime has inertial reference frames, but the typical Lortentz interpretation says only one of those reference frames is preferred: the one where the aetherwind doesn’t blow. Comment. Yes, if u are stationary in the preferred or absolute reference frame then the wind is zero kmps. In my thort-X when i say that O is stationary i don’t mean that O's apparent wind is zero, i mean O is stationary in relation to something else, but it might be a good idea to consider O to be stationary in the absolute aether frame (it might make the thort-X simpler).

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In the aether universe the present instant of time is universal

Somehow I guessed that you would also adopt this divergence from Lorentz’s ideas. Comment. I don’t remember exactly what Lorentz thort (it was so long ago). Lorentz i think thort of time in his time-gamma-stuff as being a convenient math-trick, not real, useful briefly & then best ignored (like sex). Einstein elevated time to the forefront. Neo-Lorentzians reckon that there is no such thing as time, or if u like there is but it is the now, the present instant, & the present instant is universal. Thusly nL's do not believe in time dilation, we believe in ticking dilation.

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Good points. This is solved by assuming that observational light travels at infinite speed.

Now you’re living in another universe. I can assume no such thing. You can disprove any bit of physics you like with this assumption that directly contradicts empirical measurements. Comment. Yes i can say-assume that observational light travels at infinite speed in my string-thort-X. No, i cannot disprove any bit of physics i like with that assumption. And whats with this "contradicts empirical measurements" krapp.

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We see the same problem in Einstein's train & lightning thortX. Einstein assumes that (1) observational light travels at infinite speed (but doesn't say so) & (2) assumes that an observer can see a ray of light (this would need an assistant with a smoke machine).

Nonsense. You’re making up fiction. The whole experiment revolves around finite but constant light speed. Nobody is observing rays. Light reaches a detector and is measured but once. Comment. Ok i had another look at Einstein's train-lightning-thort-X & u are correct. I was thinking of the modern youtube cartoons of this thort-X, where the platform observer is standing well away from the train. Einstein did indeed hav the 2 flashes & the 2 (midpoint) observers more or less on the same alignment (or within inches) where their eyes were directly hit by the tips of the rays (no smoke or infinite speed needed). However there was no detector, they did observe rays with their eyes (not important). In my string-thort-X i too can place my observer O within inches of the alignment of X~Y, but then the thort-X wouldn’t work proper, O needs to be well back (a very long way back), for my thort-X to make much sense. So, i need an infinite observational-light speed for O (not a problem).

Your theory has the aetherwind changing? Thatís not the neo-Lorentzian view. The background aetherwind blowing at 500 kmps south to north at 20 deg off Earth's spin-axis might not change much, but the apparent wind blowing throo u & me gradually changes during each day & year due to Earth's 0.4 kmps spin & 30 kmps orbit.

Thatís us changing velocity, not the aether.

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For example Demjanov's 1970 MMX in Obninsk showed that the horizontal projection of the wind was 140 kmps at one time of day & had a max of 480 kmps at another time of day

How the heck did he get a variance of 340 kmps if the variance of our velocity is only 61 kmps? What exactly was being measured, since the aether is not detectable? I find no references to this experiment.

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In which case the string never stretches or slackens & never breaks.

If anybody says any of that, they donít know their physics. If you want to call them Einsteinians, fine, but you seem to be talking about uneducated people.I was referring to the rezults of a poll at CERN Ė some said that the string doesnít break, some said that it breaks (last i heard all of them were still getting paid).

A poll was taken at CERN about your string question here? And they answered that way? Love to see a reference that confirms that. I personally find that very hard to believe.

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I donít know about that tall building stuff, my thort-X is out in space well away from mass etc.

More to the point: Two ships, not two ends of one long ship. It makes a very big difference.

You say this:

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Fast things might be contracted or might not be.

and then this:

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Yes, if u are stationary in the preferred or absolute reference frame then the wind is zero kmps.

The first comment implies that something might be a Ďfast thingí that is nevertheless not contracted because it is stationary in the preferred frame. That doesnít sound like a Ďfast thingí to me.Itís fine in the Einstein view where that thing is just fast is some other frame, but thatís not the typical language adopted by a self-proclaimed aetherist.

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In the aether universe the present instant of time is universal

Somehow I guessed that you would also adopt this divergence from Lorentzís ideas.I donít remember exactly what Lorentz thort (it was so long ago).

Lorentz envisioned 4D spacetime with a preferred frame but not a preferred moment. Both Minkowski and Einstein borrowed from Lorentzís work in this area in the formulation of their theories. Spacetime has no Ďpresent instantí, but yes, I find that typical nLís posit one anyway.

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Yes i can say-assume that observational light travels at infinite speed in my string-thort-X.

You are on your own then. I cannot take your examples seriously with this assumption.Is it really so hard to say that some event happens 1 light-hour away, so it gets seen 1 hour after it happens? Weíre both quite capable of subtracting off the signal travel time.

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In my string-thort-X i too can place my observer O within inches of the alignment of X~Y, but then the thort-X wouldnít work proper, O needs to be well back (a very long way back), for my thort-X to make much sense. So, i need an infinite observational-light speed. Thatís not a problem.

That works quite nicely as well, putting your observer way back, so you see all events in the order they happen in the frame of that observer. Yes, you still need to select a frame for the observer, even if heís well back. One of them can be on another train running in parallel with the first.

Not sure what this distant observer expects to see. Light signals from the two flashes going sideways to the observer? That works. He cannot see light move to some observer on the platform, but he can see the platform (or the train guy) signal some way when each flash is perceived.

Just guessing. Iíve not seen your Ďstring-thort-Xí. It isnít going to predict anything different than relativity theory.

Your theory has the aetherwind changing? That’s not the neo-Lorentzian view. Comment. If u change velocity then the apparent aetherwind blowing throo u must change, the background wind can be considered constant. The background aetherwind blowing at 500 kmps south to north at 20 deg off Earth's spin-axis might not change much, but the apparent wind blowing throo u & me gradually changes during each day & year due to Earth's 0.4 kmps spin & 30 kmps orbit.

That’s us changing velocity, not the aether. Comment. Yes, the background aetherwind probly doesnt change much during a day.

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For example Demjanov's 1970 MMX in Obninsk showed that the horizontal projection of the wind was 140 kmps at one time of day & had a max of 480 kmps at another time of day

How the heck did he get a variance of 340 kmps if the variance of our velocity is only 61 kmps? What exactly was being measured, since the aether is not detectable? I find no references to this experiment. Comment. google -- Michelson-type interferometer operating at effects of first order with respect to v/c -- V V Demjanov -- 2010.I think that the 61 kmps has only a small effect. The main difference arises due to the latitude, ie its effect on the horizontal angle of the 500 kmps. At the poles an MMX would show very little change in the horizontal component of the wind speed (praps 61 kmps), at the equator it would be at a max (praps 500 kmps), at the latitude of Obninsk (near Moscow) it was 340 kmps (on 22 June 1970).

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In which case the string never stretches or slackens & never breaks.

If anybody says any of that, they don’t know their physics. If you want to call them Einsteinians, fine, but you seem to be talking about uneducated people.I was referring to the rezults of a poll at CERN – some said that the string doesn’t break, some said that it breaks (last i heard all of them were still getting paid).

A poll was taken at CERN about your string question here? And they answered that way? Love to see a reference that confirms that. I personally find that very hard to believe. Comment. I will have a look for it.

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I don’t know about that tall building stuff, my thort-X is out in space well away from mass etc.

More to the point: Two ships, not two ends of one long ship. It makes a very big difference.You say this:

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Fast things might be contracted or might not be.

and then this:

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Yes, if u are stationary in the preferred or absolute reference frame then the wind is zero kmps.

The first comment implies that something might be a ‘fast thing’ that is nevertheless not contracted because it is stationary in the preferred frame. That doesn’t sound like a ‘fast thing’ to me.It’s fine in the Einstein view where that thing is just fast is some other frame, but that’s not the typical language adopted by a self-proclaimed aetherist. Comment. If the wind is a tailwind then a fast thing will be longer.

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In the aether universe the present instant of time is universal

Somehow I guessed that you would also adopt this divergence from Lorentz’s ideas.I don’t remember exactly what Lorentz thort (it was so long ago).

Lorentz envisioned 4D spacetime with a preferred frame but not a preferred moment. Both Minkowski and Einstein borrowed from Lorentz’s work in this area in the formulation of their theories. Spacetime has no ‘present instant’, but yes, I find that typical nL’s posit one anyway. Comment. I am dubious re Lorentz using the word spacetime before Einstein & Co. I will have a look.

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Yes i can say-assume that observational light travels at infinite speed in my string-thort-X.

You are on your own then. I cannot take your examples seriously with this assumption.Is it really so hard to say that some event happens 1 light-hour away, so it gets seen 1 hour after it happens? We’re both quite capable of subtracting off the signal travel time. Comment. Yes it is much too hard. Infinite observational speed of light fixes that.

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In my string-thort-X i too can place my observer O within inches of the alignment of X~Y, but then the thort-X wouldn’t work proper, O needs to be well back (a very long way back), for my thort-X to make much sense. So, i need an infinite observational-light speed. That’s not a problem.

That works quite nicely as well, putting your observer way back, so you see all events in the order they happen in the frame of that observer. Yes, you still need to select a frame for the observer, even if he’s well back. One of them can be on another train running in parallel with the first. Comment. The Einsteinian universe needs to consider the frame of O, the aetheric universe doesnt need to. The aetheric universe can ignore O, it only needs to consider the aetherwind blowing throo XY, this will tell whether the string stretches or slackens. Then re what does O see, O sees exactly what happens, always, no matter what is O's speed or acceleration or position (constrained only by hisher ability to see instantaneously)(which is why i prefer an infinite observational light speed).

Not sure what this distant observer expects to see. Light signals from the two flashes going sideways to the observer? That works. He cannot see light move to some observer on the platform, but he can see the platform (or the train guy) signal some way when each flash is perceived. Comment.

Just guessing. I’ve not seen your ‘string-thort-X’. It isn’t going to predict anything different than relativity theory. Comment. My string-thort-X is merely the A123 & B123 question stuff. But i know that there are versions of a spaceship-string-spaceship thort-X out there on google & wiki. I must have a look to see what they reckon.

For example Demjanov's 1970 MMX in Obninsk showed that the horizontal projection of the wind was 140 kmps at one time of day & had a max of 480 kmps at another time of day

OK, the measurements are just done at different angles, with a device too cumbersome to just point different ways, so it uses Earth rotation to do that work. I cannot figure what it is measuring. It isn't speed relative to CMB since that cannot produce a value greater than about 370 km/sec (pointed at say Leo constellation), which is the speed typically promoted by aetherists.

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If the wind is a tailwind then a fast thing will be longer.

How can something that is stationary be a tail wind?

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I am dubious re Lorentz using the word spacetime before Einstein & Co. I will have a look.

I think Minkowski coined that term. But Lorentz first envisioned it, and did the coordinate rotation mathematics for it, which is still called the Lorentz transformation.

For example Demjanov's 1970 MMX in Obninsk showed that the horizontal projection of the wind was 140 kmps at one time of day & had a max of 480 kmps at another time of day

OK, the measurements are just done at different angles, with a device too cumbersome to just point different ways, so it uses Earth rotation to do that work. I cannot figure what it is measuring. It isn't speed relative to CMB since that cannot produce a value greater than about 370 km/sec (pointed at say Leo constellation), which is the speed typically promoted by aetherists. Comment. Demjanov used the smallest MMX ever, smaller than all of the silly little Year-1 MMX's done every year in Physics 101 at Colleges all over theusofa. His MMX was small because it was a twin media (air~carbondisulphide) MMX, which was 1000 times as sensitive as the old fashioned MMXs. It certainly wasnt one of those big MMXs that were frozen in place & needed the Earth's spin to work (i think there were 2 of these somewhere, Munera had one). Demjanov's MMX spun & gave a result for fringeshift in less than a minute -- the result changing during the course of a day. Why are we even arguing about Einstein -- what sort of proof is needed that aether & aetherwind exists.

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If the wind is a tailwind then a fast thing will be longer.

How can something that is stationary be a tail wind? Comment. I didnt stipulate what the aethewind is or isnt, ie speed & direction, thusly it could be a tailwind or it could be a headwind. But stationary would be simpler for sure.

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I am dubious re Lorentz using the word spacetime before Einstein & Co. I will have a look.

I think Minkowski coined that term. But Lorentz first envisioned it, and did the coordinate rotation mathematics for it, which is still called the Lorentz transformation. Comment. Not forgetting that Lorentz & Michelson & Miller & Morley & Ives & Poincare & Silberstein all died believing in aether. And Einstein died believing in aether.

Comment. I didnt stipulate what the aethewind is or isnt, ie speed & direction, thusly it could be a tailwind or it could be a headwind. But stationary would be simpler for sure.

Yes you did:

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Yes, if u are stationary in the preferred or absolute reference frame then the wind is zero kmps.

That is pretty much an assertion that the aether is stationary relative to the absolute frame, consistent with the neo-Lorentzian view. As such, it cannot be a tail wind. If there is wind, it is because you are moving and the wind will come from your direction of movement: a head wind.

Some relativity question(s). I am especially interested in what Einstein would say, but there are other theories.

Spaceship X is connected to spaceship Y (line ahead) by a tight elastic thread & spaceship O is nearby. X & Y & O are stationary. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....(A1) X & Y accelerate at the same rate & there is no observer on O which is stationary.(A2) X & Y accelerate at the same rate & there is an observer on O which is stationary.(A3) X & Y accelerate at the same rate & there is an observer on O which accelerates likewise.

I'm not interested in the observers as the thread either breaks for all of them or doesn't break for any of them.

As X and Y accelerate, they contract in length but stay the same distance apart, so the thread snaps.

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O is stationary & X & Y are going past at hi speed connected by the tight thread. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....(B1) X & Y decelerate & there is no observer on O which is stationary.(B2) X & Y decelerate & there is an observer on O which is stationary.(B3) X & Y decelerate & there is an observer on O which accelerates at the same rate in the opposite direction.

As X and Y decelerate, they lose the length contraction that they started with, so the new thread goes slack. However, that only happens if their clocks are still synchronised for the first experiment. If they resynchronise their clocks for the frame in which they start this deceleration, the new thread will break as before because the leading ship will start its deceleration late.

Comment. I didnt stipulate what the aethewind is or isnt, ie speed & direction, thusly it could be a tailwind or it could be a headwind. But stationary would be simpler for sure.

Yes you did: Comment. No, not in the original questions. But the aetheric answer demands the knowledge of the initial background aetherwind (which at the same time tells u about the initial apparent wind)(which at the same time tells u about the change in the apparent wind during the events), & depending on the initial wind the aetheric answer can be (a) stretches, or (b) slackens, or (c) stretches then slackens, or (d) slackens then stretches -- but with the knowledge of the initial wind the answer in each of A123 & B123 will be one of these choices, & the answers for each of the 6 cases might be different, except that the answer for A1 will allways be the same as for A2, & B1 will allways be the same as B2 (unless u nominate different winds).

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Yes, if u are stationary in the preferred or absolute reference frame then the wind is zero kmps.

That is pretty much an assertion that the aether is stationary relative to the absolute frame, consistent with the neo-Lorentzian view. As such, it cannot be a tail wind. If there is wind, it is because you are moving and the wind will come from your direction of movement: a head wind. Comment. It can be a tailwind, depending on what u nominate, & as i said above the aetheric answer depends on the kmps & degrees nominated for the wind (whereas the Einsteinian answer probably doesnt need any such nomination).

Some relativity question(s). I am especially interested in what Einstein would say, but there are other theories.

Spaceship X is connected to spaceship Y (line ahead) by a tight elastic thread & spaceship O is nearby. X & Y & O are stationary. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....(A1) X & Y accelerate at the same rate & there is no observer on O which is stationary.(A2) X & Y accelerate at the same rate & there is an observer on O which is stationary.(A3) X & Y accelerate at the same rate & there is an observer on O which accelerates likewise.

I'm not interested in the observers as the thread either breaks for all of them or doesn't break for any of them. Comment. That is of course in an Einsteinian universe (& is likewise the the aetheric universe)(depending on the initial wind nominated)

As X and Y accelerate, they contract in length but stay the same distance apart, so the thread snaps. Comment. My Einsteinian answer is that the thread doesnt stretch or slacken or snap. But i think i noticed in wiki that the knee-jerk Einsteinian answer is that there is no change -- & wiki says that a more considered Einsteinian answer needs to account for simultaneity, in which case the thread snaps. I will have to look into that. If your answer refers to the aetheric universe then as i have said elsewhere u need to know the initial wind before answering. But i suppose that u could say that your answer is based on what happens in the long run, ie after a long time -- but here we might get into trouble because in the long run the ultimate speed will be nearly c (on the other hand perhaps there aint a problem here)(i think there aint a problem). But anyhow re such possible trouble -- in the aetheric universe it is possible for X~Y to be moving at almost c in one direction, & for O to be moving at almost c in the opposite direction, in which case the relative speed is almost 2c, however light from X~Y will eventually reach O after a long time (which makes my stipulation for an infinite speed of observational light even more necessary).

Re that there relative speed of almost 2c -- here i mean true relative speed, ie as observed by an observer in the absolute aether rest frame (where the aetherwind for the observer is 00 kmps). For our observer O, who might have an aetherwind blowing throo himher, hisher eyes will be deformed (contracted) by the wind, & heshe will always see the true lengths & true speeds of X & string & Y. Likewise observers X & Y will usually have deformed eyes & they too will always see the true lengths & speeds of X & string & Y (so i don know why i needed to write this paragraph)(ignore it).

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O is stationary & X & Y are going past at hi speed connected by the tight thread. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....(B1) X & Y decelerate & there is no observer on O which is stationary.(B2) X & Y decelerate & there is an observer on O which is stationary.(B3) X & Y decelerate & there is an observer on O which accelerates at the same rate in the opposite direction.

As X and Y decelerate, they lose the length contraction that they started with, so the new thread goes slack. However, that only happens if their clocks are still synchronized for the first experiment. If they resynchronize their clocks for the frame in which they start this deceleration, the new thread will break as before because the leading ship will start its deceleration late. Comment. I dont understand how there can ever be a simultaneity~synchronization issue for X & Y in A123 & B123 (i mean assuming that they successfully synchronize at some time before)(& assuming that they do the same thing at the same time up untill the event)(& during the event). During the event they always do exactly the same thing (accelerate or decelerate) at exactly the same time. So why the problem. Am i missing something. Especially as all of this happens in deep outer space a long way from mass (such mass might affect the clocks differentially if nearer to one).How can the leading ship X start its acceleration late?

In the aetheric universe the clocks X & Y would too have to be synchronized at the start of the event, & here too i dont see how they would lose synchronicity if they both did exactly the same things thereafter. In B3 clock O would have to be synchronized with X & Y.

Note that in the original question you can find the wording "are stationary". That already implies that there is a fabric of space which they are stationary relative to, so it invites answers like mine which are based on Lorentz Ether Theory rather than SR.

My Einsteinian answer is that the thread doesnt stretch or slacken or snap. But i think i noticed in wiki that the knee-jerk Einsteinian answer is that there is no change -- & wiki says that a more considered Einsteinian answer needs to account for simultaneity, in which case the thread snaps. I will have to look into that.

The correct Einsteinian answers match the correct LET answers, other than in the mechanism (because they accept contradictory accounts as all being equally correct, whereas LET recognises only one account as accurate due to its more rational rejection of contradiction).

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I dont understand how there can ever be a simultaneity~synchronization issue for X & Y in A123 & B123 (i mean assuming that they successfully synchronize at some time before)(& assuming that they do the same thing at the same time up untill the event)(& during the event). During the event they always do exactly the same thing (accelerate or decelerate) at exactly the same time. So why the problem. Am i missing something.

No, you aren't missing anything - so long as they stick with the original synchronisation, every simultaneous manoeuvre that they make will be simultaneous in the original frame, so the thread will snap on acceleration, then the replacement one will go slack as they decelerate.

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How can the leading ship X start its acceleration late?

If they resynchronise their clocks after accelerating and do so on the basis that they're now stationary, they'll end up with the leading one running behind the other (when observed by a stationary observer in the original frame). That's why the leading one will start decelerating after the trailing one, but they will see themselves as simultaneously beginning that deceleration. If they don't resynchronise their clocks though, the leading ship will not start the deceleration late, and to the people in the two ships it will appear as if the leading one starts its deceleration first.

Note that in the original question you can find the wording "are stationary". That already implies that there is a fabric of space which they are stationary relative to, so it invites answers like mine which are based on Lorentz Ether Theory rather than SR.

My Einsteinian answer is that the thread doesnt stretch or slacken or snap. But i think i noticed in wiki that the knee-jerk Einsteinian answer is that there is no change -- & wiki says that a more considered Einsteinian answer needs to account for simultaneity, in which case the thread snaps. I will have to look into that.

The correct Einsteinian answers match the correct LET answers, other than in the mechanism (because they accept contradictory accounts as all being equally correct, whereas LET recognises only one account as accurate due to its more rational rejection of contradiction).

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I dont understand how there can ever be a simultaneity~synchronization issue for X & Y in A123 & B123 (i mean assuming that they successfully synchronize at some time before)(& assuming that they do the same thing at the same time up untill the event)(& during the event). During the event they always do exactly the same thing (accelerate or decelerate) at exactly the same time. So why the problem. Am i missing something.

No, you aren't missing anything - so long as they stick with the original synchronisation, every simultaneous manoeuvre that they make will be simultaneous in the original frame, so the thread will snap on acceleration, then the replacement one will go slack as they decelerate. Comment. No i dont see the thread snapping. Unless u mean from the g force of acceleration.

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How can the leading ship X start its acceleration late?

If they resynchronise their clocks after accelerating and do so on the basis that they're now stationary, they'll end up with the leading one running behind the other (when observed by a stationary observer in the original frame). That's why the leading one will start decelerating after the trailing one, but they will see themselves as simultaneously beginning that deceleration. If they don't resynchronise their clocks though, the leading ship will not start the deceleration late, and to the people in the two ships it will appear as if the leading one starts its deceleration first. Comment. Wow, if thats the Einsteinian answer then Einsteinians are crazier than i ever thort. I dont see how X or Y can ever see the thread snap in the Einsteinian universe -- X & Y must always think themselves to be stationary relative to each other.And i dont see how O might see the thread snap. I get it that Einsteinian SR says that if X & Y are simultaneous in their frame then they cannot be simultaneous in any other frame (this rule is wrong)(but i can follow their logic)(albeit flawed). Thusly O will see a difference in activity of X compared to Y, but the thread doesnt snap in the XY frame, therefore the thread doesnt snap, therefore O cannot see the thread snap (still sticking to Einstein's SR rules in an Einstein universe). Anything else is batshit crazy.

Comment. No i dont see the thread snapping. Unless u mean from the g force of acceleration.

Length contraction of the ships will cause the thread to break on acceleration. The ships maintain the same distance between the centres of mass of each, but because they both contract in length, the thread will snap. This happens both in LET (Lorentz Ether Theory) and in SR.

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If they resynchronise their clocks after accelerating and do so on the basis that they're now stationary, they'll end up with the leading one running behind the other (when observed by a stationary observer in the original frame). That's why the leading one will start decelerating after the trailing one, but they will see themselves as simultaneously beginning that deceleration. If they don't resynchronise their clocks though, the leading ship will not start the deceleration late, and to the people in the two ships it will appear as if the leading one starts its deceleration first.

Comment. Wow, if thats the Einsteinian answer then Einsteinians are crazier than i ever thort. I dont see how X or Y can ever see the thread snap in the Einsteinian universe -- X & Y must always think themselves to be stationary relative to each other.

They are stationary relative to each other, but you make a good point - as they accelerate, the lead one will appear to accelerate faster than the trailing one from the point of view of the people in those two ships because the apparent distance between them will increase. The leading one will also appear to stop accelerating sooner while the trailing one appears to continue accelerating until it's up to the same speed, and yet both of them accelerated at the same rate for the same length of time. When they decelerate, if they do this without resynchronising clocks, the trailing one will appear to start its deceleration first (from the point of view of the people in those two ships) and then the leading one will appear to decelerate more strongly, and yet again the reality is that they both started their deceleration at the same time and decelerated at the same rate. This is again an LET description, but it takes you straight to the correct SR description too - it's much harder to work out what would happen if you only know SR because it's so easy to misunderstand the rules in SR. Your idea that the ships should appear to remain stationary relative to each other under acceleration from the point of view of the occupants in SR is incorrect.

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And i dont see how O might see the thread snap.

The thread either snaps or it doesn't. If it snaps and O can't see it, O needs to buy a bigger telescope.

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I get it that Einsteinian SR says that if X & Y are simultaneous in their frame then they cannot be simultaneous in any other frame (this rule is wrong)(but i can follow their logic)(albeit flawed). Thusly O will see a difference in activity of X compared to Y, but the thread doesnt snap in the XY frame, therefore the thread doesnt snap, therefore O cannot see the thread snap (still sticking to Einstein's SR rules in an Einstein universe). Anything else is batshit crazy. [/color]

There is no question of the thread snapping for some observers and not for others - no theory of relativity allows that. The thread will always snap if the acceleration starts at the same time for leading and following ship with clocks synchronised for the frame in which they begin from at rest. If they synchronise the clocks in advance for the target frame in which they'll be at rest once they've stopped accelerating, then the thread will go slack instead. If they start with the clocks synchronised for a frame half way in between and start with a slack line with just the right amount of slack in it, it could become tight half way through and then go slack again. This applies equally for LET and SR.